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Gọi \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{31}\) là S
Ta có:
\(S=1+\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}\right)+\left(\dfrac{1}{8}+\dfrac{1}{9}+...+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+\dfrac{1}{17}+...+\dfrac{1}{31}\right)\)
\(S< 1+\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}+...+\dfrac{1}{8}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}+...+\dfrac{1}{16}\right)\)
\(S< 1+1+1+1+1\)
\(S< 5\)
Vậy \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{31}< 5\)
Đặt \(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}=A\)
Ta có : \(A=\frac{1}{5}+\left(\frac{1}{14}+\frac{1}{28}+\frac{1}{44}\right)+\left(\frac{1}{61}+\frac{1}{85}+\frac{1}{97}\right)\)
\(A< \frac{1}{5}\left(\frac{1}{14.3}\right)+\left(\frac{1}{61.3}\right)\)
\(A< \frac{1}{5}+\frac{3}{14}+\frac{3}{61}\)
\(A< \frac{1}{5}+\frac{3}{12}+\frac{1}{20}\)
\(A< \frac{1}{2}\left(ĐPCM\right).\)
\(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}< \frac{1}{2}\)
\(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}=0,36833......\)
mà \(\frac{1}{2}=0,5\)
\(0,36833..< 0,5\)
Vậy \(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}< \frac{1}{2}\)
Đặt \(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}=A\)
Ta có : \(A=\frac{1}{5}+\left(\frac{1}{14}+\frac{1}{28}+\frac{1}{44}\right)+\left(\frac{1}{61}+\frac{1}{85}+\frac{1}{97}\right)\)
\(A< \frac{1}{5}\left(\frac{1}{14.3}\right)+\left(\frac{1}{61.3}\right)\)
\(A< \frac{1}{5}+\frac{3}{14}+\frac{3}{61}\)
\(A< \frac{1}{5}+\frac{3}{12}+\frac{1}{20}\)
\(A< \frac{1}{2}\left(ĐPM\right)\).
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)
Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ; (1/51 + 1/52+...+1/59+1/60) > 1/6
S > 1/4 + 1/5 + 1/6.
Trong khi đó (1/4 + 1/5 + 1/6) > 3/5
=>S > 3/5 (1)
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)
=> S < 4/5 (2)
Từ (1) và (2) => 3/5 <S<4/5
Ta có: \(\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}.\)
\(=\frac{1}{5}+\left(\frac{1}{14}+\frac{1}{31}+\frac{1}{44}\right)+\left(\frac{1}{61}+\frac{1}{84}+\frac{1}{96}\right)\)
Ta thấy \(\frac{1}{14}< \frac{1}{12}\)
\(\frac{1}{31}< \frac{1}{12}\)
\(\frac{1}{44}< \frac{1}{12}\)
\(=>\frac{1}{14}+\frac{1}{31}+\frac{1}{44}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}\)
\(=>\frac{1}{14}+\frac{1}{31}+\frac{1}{44}< \frac{1}{12}.3\left(1\right)\)
Ta lại thấy \(\frac{1}{61}< \frac{1}{60}\)
\(\frac{1}{84}< \frac{1}{60}\)
\(\frac{1}{96}< \frac{1}{60}\)
\(=>\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}\)
\(=>\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{60}.3\left(2\right)\)
Từ (1) và (2) suy ra: \(\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3\)
\(=>\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{5}+3.\left(\frac{1}{12}+\frac{1}{60}\right)\)
\(=>\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{2}\)
\(=>Đpcm\)